What are the solutions to the equation (1-x)(11-x)+26=0 ?

The solutions to the equation (1-x)(11-x)+26=0 are x=6+i,x=6-i

Find the zeros of (1-x)(11-x)+26=0

The zeros of (1-x)(11-x)+26=0 are x=6+i,x=6-i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(1 - x) (11 - x) + 26 = 0

x = 6 + i, x = 6 - i

Solution steps

(1 - x) (11 - x) + 26 = 0

Expand

(1 - x) (11 - x) + 26 : x^{2} - 12 x + 37

x^{2} - 12 x + 37 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 37}{2 \cdot 1}

Simplify

(- 12)^{2} - 4 \cdot 1 \cdot 37 : 2 i

x_{1, 2} = \frac{- ( - 12 ) \pm 2 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 i}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 2 i}{2 \cdot 1}

x = \frac{- ( - 12 ) + 2 i}{2 \cdot 1} : 6 + i

x = \frac{- ( - 12 ) - 2 i}{2 \cdot 1} : 6 - i

The solutions to the quadratic equation are:

x = 6 + i, x = 6 - i

8 x^{2} + 3 = 2

x^{2} + 14 x + 6 = 0

x^{2} + x = - 5

2 t^{2} + t + 9 = 2

s^{2} + 4 s + 29

What are the solutions to the equation (1-x)(11-x)+26=0 ?
The solutions to the equation (1-x)(11-x)+26=0 are x=6+i,x=6-i
Find the zeros of (1-x)(11-x)+26=0
The zeros of (1-x)(11-x)+26=0 are x=6+i,x=6-i